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    (Original post by Renzhi10122)
    Problem 583

     x_i=\pm 1 and  x_1x_2x_3x_4+x_2x_3x_4x_5+...+x_  nx_1x_2x_3=0
    Prove that  n is divisible by 4.
    A beauty, only invariant question I got off that problem sheet. I was so pleased.
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    (Original post by physicsmaths)
    A beauty, only invariant question I got off that problem sheet. I was so pleased.
    Lol, I quite like it.
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    (Original post by Renzhi10122)
    Problem 583

     x_i=\pm 1 and  x_1x_2x_3x_4+x_2x_3x_4x_5+...+x_  nx_1x_2x_3=0
    Prove that  n is divisible by 4.
    Solution 583:
    Method 1:
    Spoiler:
    Show
    Let y_i=x_ix_{i+1}x_{i+2}x_{i+3} with x_{n+j} \rightarrow x_j for j \in \{1,2,3\}.
    Then \displaystyle\sum_{i=1}^n y_i = 0 \iff 2|n, so y_{i_r}=-1 for some i_r with r \in \{1, . .,\frac{n}{2} \}.
    Note also that:
    \displaystyle\prod_{r=1}^{n/2} y_{i_r} =\prod_{i=1}^n y_i = \prod_{i=1}^n x_i^4 = 1 \implies (-1)^{n/2}=1 \iff 4|n.
    Method 2:
    Spoiler:
    Show
    Write S=x_1x_2x_3x_4+. . .+x_nx_1x_2x_3.
    Note that S = 0, so S \equiv 0 \pmod{4}.
    Note that if we change the sign of any of the x_i then we change 4 terms of S by either \pm 2.
    Regardless of any changes to the terms, we still have S \equiv 0 \pmod{4}.
    Now note that if each x_i=1, then S=n, and so S \equiv 0 \pmod{4} \implies n \equiv 0 \pmod{4}.
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    (Original post by joostan)
    Solution 583:
    Method 1:
    Spoiler:
    Show
    Let y_i=x_ix_{i+1}x_{i+2}x_{i+3} with x_{n+j} \rightarrow x_j for j \in \{1,2,3\}.
    Then \displaystyle\sum_{i=1}^n y_i = 0 \iff 2|n, so y_{i_r}=-1 for some i_r with r \in \{1, . .,\frac{n}{2} \}.
    Note also that:
    \displaystyle\prod_{r=1}^{n/2} y_{i_r} =\prod_{i=1}^n y_i = \prod_{i=1}^n x_i^4 = 1 \implies (-1)^{n/2}=1 \iff 4|n.
    Method 2:
    Spoiler:
    Show
    Write S=x_1x_2x_3x_4+. . .+x_nx_1x_2x_3.
    Note that S = 0, so S \equiv 0 \pmod{4}.
    Note that if we change the sign of any of the x_i then we change 4 terms of S by either \pm 2.
    Regardless of any changes to the terms, we still have S \equiv 0 \pmod{4}.
    Now note that if each x_i=1, then S=n, and so S \equiv 0 \pmod{4} \implies n \equiv 0 \pmod{4}.
    Yup, method 2 was the one I knew, but I quite like method 1.
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    (Original post by joostan)
    Solution 583:
    Method 1:
    Spoiler:
    Show
    Let y_i=x_ix_{i+1}x_{i+2}x_{i+3} with x_{n+j} \rightarrow x_j for j \in \{1,2,3\}.
    Then \displaystyle\sum_{i=1}^n y_i = 0 \iff 2|n, so y_{i_r}=-1 for some i_r with r \in \{1, . .,\frac{n}{2} \}.
    Note also that:
    \displaystyle\prod_{r=1}^{n/2} y_{i_r} =\prod_{i=1}^n y_i = \prod_{i=1}^n x_i^4 = 1 \implies (-1)^{n/2}=1 \iff 4|n.
    Method 2:
    Spoiler:
    Show
    Write S=x_1x_2x_3x_4+. . .+x_nx_1x_2x_3.
    Note that S = 0, so S \equiv 0 \pmod{4}.
    Note that if we change the sign of any of the x_i then we change 4 terms of S by either \pm 2.
    Regardless of any changes to the terms, we still have S \equiv 0 \pmod{4}.
    Now note that if each x_i=1, then S=n, and so S \equiv 0 \pmod{4} \implies n \equiv 0 \pmod{4}.
    Yep, this is from a problem sheet on invariants so yeh changibg the 1s to -1s etc is a nice way but Method 1 is very nice aswell.
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    This problem is for any younger lurkers, as it only requires basic trig, calculus and geometry. I'd post it on the STEP thread, but they have better things to do than do what i spent the better part of a few hour messing around with

    Problem 584:

    A clean rainbow appears due to an object of sufficient symmetry scattering light from a source. Such light will appear at a certain angle along the line of the observer and the light source due to a focusing effect, such that a complete circle would form (if the ground isn't in the way). Let the order of a rainbow be related to the number of reflections inside such an object.

    Let the light source be the sun and the scattering object be perfectly spherical raindrops (refractive index = 4/3). Explain mathematically why the focusing of light occurs at a particular angle for primary rainbows. Why would secondary rainbows look so different to primary rainbows? Find a general solution to angles which rainbows can exist along the line connecting the sun to the observer due to such focusing. Why do primary rainbows not have the expected width of 1.9 degrees, but rather one closer to 2.5 degrees?

    Violet light has a refractive index of ~1.344
    The sun is sufficiently far away that incident rays are parallel

    Rough diagram of what is going on in a primary rainbow for anyone completely lost:
    Spoiler:
    Show
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    Problem 585**

     \sum^n_{i=1}x_i=-n, \sum^n_{i=1}x_i^{32}=n, x_i \in \mathbb{R} . Prove that  x_i=-1 .
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    (Original post by Renzhi10122)
    Problem 585**

     \sum^n_{i=1}x_i=-n, \sum^n_{i=1}x_i^{32}=n, x_i \in \mathbb{R} . Prove that  x_i=-1 .
    failed attempt.
    Ew :lol: think this is okay? Not ever good at proofs :P
    Spoiler:
    Show
    Note that \sum^n_{i=1}x_i = nx.
    \sum^n_{i=1}x_i^{32}=n, x_i = nx^{32}

    nx=-n, nx^{32}

    x=-\frac{n}{n} = -1 and

    nx=nx^{32} \implies x=x^{32} \implies x=\pm 1

    Only -1 satisfies both arguments, therefore
    x=-1
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    (Original post by The-Spartan)
    Solution 585
    Ew :lol: think this is okay? Not ever good at proofs :P
    Spoiler:
    Show
    Note that \sum^n_{i=1}x_i = nx.
    \sum^n_{i=1}x_i^{32}=n, x_i = nx^{32}

    nx=-n, nx^{32}

    x=-\frac{n}{n} = -1 and

    nx=nx^{32} \implies x=x^{32} \implies x=\pm 1

    Only -1 satisfies both arguments, therefore
    x=-1
    You seem to have assumed each x_i is the same. . .
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    (Original post by The-Spartan)
    Solution 585
    Ew :lol: think this is okay? Not ever good at proofs :P
    Spoiler:
    Show
    Note that \sum^n_{i=1}x_i = nx.
    \sum^n_{i=1}x_i^{32}=n, x_i = nx^{32}

    nx=-n, nx^{32}

    x=-\frac{n}{n} = -1 and

    nx=nx^{32} \implies x=x^{32} \implies x=\pm 1

    Only -1 satisfies both arguments, therefore
    x=-1
    (Original post by joostan)
    You seem to have assumed each x_i is the same. . .
    Ah, I was wondering what was happening... Yeah, it's a bit more complicated than that.
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    (Original post by Renzhi10122)
    Ah, I was wondering what was happening... Yeah, it's a bit more complicated than that.
    Damn :lol:
    Didnt notice that assumption. Was wondering why it was a ** label
    Thanks joostan
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    Problem 583

    For anyone as fed up of studying as I am. Evaluate

    (n+1) + 2n + 2^2 (n - 1) + ... + 2^k (n + 1 - k) + .... + 2^n
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    (Original post by 16Characters....)
    Problem 583

    For anyone as fed up of studying as I am. Evaluate

    (n+1) + 2n + 2^2 (n - 1) + ... + 2^k (n + 1 - k) + .... + 2^n
    Why not, note:

    \displaystyle\sum_{k=0}^{n} x^k = \dfrac{1-x^{n+1}}{1-x}.

    Differentiating this w.r.t. x, also have:

    \displaystyle\sum_{k=0}^{n} kx^{k-1} = \dfrac{1-(n+1)x^n +nx^{n+1}}{(1-x)^2}

    So,

    

\begin{array}{rcl}

\displaystyle\sum_{k=0}^{n} 2^k(n+1-k) &=& (n+1)\displaystyle\sum_{k=0}^{n} 2^k - 2\displaystyle\sum_{k=0}^{n} k2^{k-1} \\

&=& (n+1)(2^{n+1}-1) - 2(1-(n+1)2^n+n2^{n+1})\\

\\

&=& 2^{n+2}-n-3

\end{array}
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    Solution 585

    Spoiler:
    Show


    The problem is equivalent to proving that for any n, having the two conditions
    x1 + ... + xn = n,
    (x1)^32 + ... + (xn)^32 = n,
    implies that all of the xi equal 1.

    Lemma: given real x1 + x2 = S, then T = (x1)^32 + (x2)^32 has minimum value 2(S/2)^32, achieved iff x1=x2 = S/2.

    Proof.
    Let x1 = S/2 + d, x2 = S/2 - d. Then T equals 2(S/2)^32 plus a load of even-exponentiated, non-negative terms which are positive (so T > 2(S/2)^32) unless d=0.

    If our xi, satisfying x1 + ... + xn = n, are not all equal, we can take any two different ones, say x1 + x2 = S, and replace them by their average (conserving the sum x1 + ... + xn = n), while decreasing the sum (x1)^32 + ... + (xn)^32 at each step. At the end of this process all the xi are equal, and since their sum is n, they all equal 1. Then the sum of their 32-exponents also equals n. So having any xi not equalling 1 would have meant that the sum of the 32-exponents was larger than n, a contradiction.



    So the 32-exponent was entirely arbitrary, a bit unsurprisingly. Any even non-zero exponent would've done the job.
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    I've just figured these results out. Not sure how well known they are, so they may be easier than I think:

    Show that:

    \displaystyle \sum_{n=1}^\infty \frac{\sin \frac{n \pi}{b}-\sin \frac{n \pi}{a}}{n} = \frac{1}{2}(\frac{\pi}{a}-\frac{\pi}{b})

    \displaystyle \sum_{n=1}^\infty \frac{\cos \frac{n \pi}{b}-\cos \frac{n \pi}{a}}{n} = \ln(\frac{\sin \frac{\pi}{2a}}{\sin \frac{\pi}{2b}})
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    (Original post by atsruser)
    \displaystyle \sum_{n=1}^\infty \frac{\sin \frac{n \pi}{b}-\sin \frac{n \pi}{a}}{n} = \frac{1}{2}(\frac{\pi}{a}-\frac{\pi}{b})

    \displaystyle \sum_{n=1}^\infty \frac{\cos \frac{n \pi}{b}-\cos \frac{n \pi}{a}}{n} = \ln(\frac{\sin \frac{\pi}{2a}}{\sin \frac{\pi}{2b}})
    The first follows from this. A parallel manipulation gives \sum v^{-1}\cos vx=-\log \left[2\sin \frac{x}{2}\right].
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    (Original post by Lord of the Flies)
    The first follows from this. A parallel manipulation gives \sum v^{-1}\cos vx=-\log \left[2\sin \frac{x}{2}\right].
    Damn, I'm behind the times, aren't I. There I was, thinking I'd found a nice, fancy result, designed to challenge the finest minds of TSR.

    It would take me quite a while to follow your derivation - looks very sophisticated. I got to it via a rather simpler route (and easier than those others posted, I think). I'll put it up later.

    Oh, and having followed that thread a bit from there - cimer? Verlan?
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    (Original post by atsruser)
    It would take me quite a while to follow your derivation - looks very sophisticated. I got to it via a rather simpler route (and easier than those others posted, I think). I'll put it up later.
    Cool - and nah it isn't too sophisticated don't worry. It looks that way because I generalised and went into some nice consequences, but the evaluation of the initial series isn't too long [starts at "solution 142" obvz].

    Oh, and having followed that thread a bit from there - cimer? Verlan?
    Yup! Ah I miss the good old days when this thread was on fire.
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    (Original post by Lord of the Flies)
    The first follows from this. A parallel manipulation gives \sum v^{-1}\cos vx=-\log \left[2\sin \frac{x}{2}\right].
    This is great!

    Though can you explain this step; not sure I quite follow (in particular the second term on the last line):

    Furthermore, observe that:

    \displaystyle\begin{aligned} \int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \sum_{v\geq 1}\frac{\sin vt_{2k}}{v}\,dt_{2k}\cdots dt_1=\frac{1}{2}\int_0^x\int_0^{  t_1}\cdots \int_0^{t_{2k-1}} \pi -t_{2k}\, dt_{2k}\cdots\, dt_1

    gives

    \displaystyle\begin{aligned} (-1)^k\sum_{v\geq 1} \frac{\sin vx}{v^{2k+1}}+\sum_{i=1}^{k} \left( \frac{(-1)^{k+i+1}x^{2(k-i)+1}}{(2(k-i)+1)!}\sum_{v\geq 1}\frac{1}{v^{2i}} \right)= \frac{x^{2k}}{2}\left(\frac{\pi}  {(2k)!}-\frac{x}{(2k+1)!}\right)
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    (Original post by Lord of the Flies)
    Cool - and nah it isn't too sophisticated don't worry. It looks that way because I generalised and went into some nice consequences, but the evaluation of the initial series isn't too long [starts at "solution 142" obvz].
    I did this: we have:

    I = \int_{\frac{\pi}{a}}^{\frac{\pi}  {b}} \frac{dx}{e^{ix}-1} = \int_{\frac{\pi}{a}   }^{ \frac{\pi}{b} } \big( -\frac{1}{2} + i \frac{\sin x}{2(\cos x-1)} \big) \ dx = \frac{1}{2}(\frac{\pi}{a} - \frac{\pi}{b}) + i \ln(\frac{\sin \frac{\pi}{2a}}{\sin \frac{\pi}{2a}})

    and:

    \displaystyle I = \int_{\frac{\pi}{a}}^{\frac{\pi}  {b}} \sum_{n=1}^{\infty} e^{-inx} \ dx = \sum_{n=1}^{\infty} \int_{\frac{\pi}{a}}^{\frac{\pi}  {b}} \big( \cos nx - i \sin nx      \big) \ dx =

    \displaystyle \sum_{n=1}^{\infty} \frac{\sin \frac{n\pi}{b} - \sin \frac{n\pi}{b}}{n} + i   \sum_{n=1}^{\infty} \frac{\cos \frac{n\pi}{b} - \cos \frac{n\pi}{b}}{n}

    (which I think is OK...)

    Yup! Ah I miss the good old days when this thread was on fire.
    Right. I was wondering if "cimer" was what people did when they got to the top of a mountain or something. Are you bilingual?

    As for the good old days, yeah, those days are gone - now you've just got to slum it with second raters like me (and Zacken, of course).
 
 
 
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